Problem 37 - Thomas Calculus

Question

Determining Convergence or Divergence In Exercises \(17-44,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$\sum_{n=1}^{\infty} \frac{n !}{(2 n+1) !}$$

Step-by-Step Solution

Answer

The series converges by the Ratio Test.

Step 1: Understanding the Series

The given series is \( \sum_{n=1}^{\infty} \frac{n!}{(2n+1)!} \). We are tasked with determining whether this series converges or diverges.

Step 2: Applying the Ratio Test

To apply the Ratio Test, we compute the ratio of consecutive terms: \( \frac{a_{n+1}}{a_n} \). Let \( a_n = \frac{n!}{(2n+1)!} \). Then, \( a_{n+1} = \frac{(n+1)!}{(2(n+1)+1)!} = \frac{(n+1)!}{(2n+3)!} \).

Step 3: Compute the Ratio

Calculate the ratio \( \frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(2n+3)!} \times \frac{(2n+1)!}{n!} \). Simplify this expression: \[ \frac{a_{n+1}}{a_n} = \frac{(n+1)}{(2n+3)(2n+2)} \].

Step 4: Assessing the Limit

Take the limit as \( n \to \infty \): \[ \lim_{n \to \infty} \frac{(n+1)}{(2n+3)(2n+2)} \]. \( (2n+3)(2n+2) \approx 4n^2 \) as \( n \to \infty \). Thus, the limit becomes \( \lim_{n \to \infty} \frac{n+1}{4n^2} = 0 \).

Step 5: Conclusion Using the Ratio Test

The limit \( \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = 0 < 1 \). Because the limit is less than 1, by the Ratio Test, the series \( \sum_{n=1}^{\infty} \frac{n!}{(2n+1)!} \) converges.

Key Concepts

Ratio Test

The Ratio Test is a powerful tool for determining whether an infinite series converges or diverges. It involves taking the limit of the ratio of consecutive terms in the series. In simpler terms, if you have a series \( a_n \), calculate the ratio \( \frac{a_{n+1}}{a_n} \) for each term. Then, find the limit of this ratio as \( n \) approaches infinity. The outcome of this limit will help us decide:

• If the limit is less than 1, the series converges.
• If the limit is greater than 1, the series diverges.
• If the limit equals 1, the test is inconclusive.

In our problem with the series \( \sum_{n=1}^{\infty} \frac{n!}{(2n+1)!} \), calculating the ratio and its limit helps us find that the series converges, as the limit is 0, which is less than 1.

Factorial Series

Factorials often appear in mathematical series and are expressed using the symbol \( ! \). The factorial of a non-negative integer \( n \), written as \( n! \), is the product of all positive integers less than or equal to \( n \). Thus, \( n! = n \times (n-1) \times (n-2) \times \, ... \, \times 1 \). Factorials grow rapidly, which can affect the convergence of a series significantly. In the series in question, \( n! \) and \( (2n+1)! \) are both factorial expressions. The term \( \frac{n!}{(2n+1)!} \) indicates that the denominator grows faster, given its factorial nature. This rapid increase in the denominator compared to the numerator influences the series to converge by making each term shrink rapidly as \( n \) increases. This shrinking behavior of terms is critical in determining series convergence using tests like the Ratio Test.

Limits in Calculus

Limits are fundamental in calculus and are used to determine the behavior of functions as they approach a specific point or infinity. In the context of series and sequences, limits help evaluate the expression as terms go to infinity, which is particularly useful for convergence tests such as the Ratio Test.To find a limit, you'll generally simplify the expression so that you can observe behavior as \( n \to \infty \). For our problem, we evaluated the limit of the ratio \( \frac{(n+1)}{(2n+3)(2n+2)} \). As \( n \) grows very large, the terms in the denominator overpower the numerator, reducing the limit towards zero. Hence, understanding limits not only assists in concluding the convergence of sequences and series but also underpins much of calculus, providing a foundation for derivatives and integrals.